A force \(\mathrm{f}=\mathrm{x}^2 \mathrm{y} \hat{\mathrm{i}}+\mathrm{y}^2 \hat{\mathrm{j}}\) acts on a particle in a plane \(\mathrm{x}+\mathrm{y}=10\). The work done by this force during a displacement from \((0,0)\) to \((4 \mathrm{~m}, 2 \mathrm{~m})\) is Joule (round off to the nearest integer)

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Correct Answer

(A) 152

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Detailed explanation

\(\begin{aligned} & y=10-x \\ & w=\int_0^4 x^2(10-x) d x+\int_0^2 y^2 d y \\ & =\frac{10 x^3}{3}-\left.\frac{x^4}{4}\right|_0 ^4+\left.\frac{y^3}{3}\right|_0 ^2 \\ & =\frac{640}{3}-\frac{256}{4}+\frac{8}{3} \\ & =216 \times 64 \\ & =152 \mathrm{~J}\end{aligned}\)

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